35. Construct a Triangle Given the Angle Opposite Its Base and the Sum and Product of the Two Other Sides

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This Demonstration shows how to construct a triangle given the angle opposite the side , the sum of the lengths of the other two sides and their product . The numbers and satisfy the quadratic equation . The problem is solved graphically using Carlyle's circle.



1. Draw a horizontal line , a point on and a vertical line through . On , draw points and on opposite sides of so that and . On , draw points and on opposite sides of so that and is parallel to . Then .

2. Construct the point on on the same side as at a distance 1 from . Construct the point at a distance from and a distance from . The circle with diameter is the Carlyle circle of the quadratic equation. The circle intersects at the points and . The lengths and are solutions of the quadratic equation.

3. Draw a ray through that forms the angle with . Let be the intersection of the ray and the circle with center and radius .

4. The triangle is the desired solution.


By step 1, .

Since is the Carlyle circle of the equation, and .

By step 3, and .


Contributed by: Izidor Hafner and Marko Razpet (September 2018)
Open content licensed under CC BY-NC-SA


This task is posed as problem 471 in [1].


[1] L. H. von Holleben and P. Gerwien, Aufgaben-Systeme und Sammlungen aus der Ebenen Geometrie: Aufgaben, Vols. I and II, Berlin: G. Reimer, 1831 and 1832.


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